Properties of inverse trigonometric functions

Since \(\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}\) and \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\). In such questions, find the trigonometric value first.

By definition, \(\sin^{-1}x\) gives the angle \(y\in[-\pi/2,\pi/2]\) such that \(\sin y=x\). Applying sine to that angle returns the original value: \(\sin(\sin^{-1}x)=x\) for \(x\in[-1,1]\). Option B (\(\sin^{-1}x\)) is incorrect because it is the inverse value itself; composing in the given order returns \(x\), not the inverse. Exam tip: always check the domain of the inner function and the principal range of the inverse to verify such compositions quickly.

The function \(\cos^{-1}x\) returns the principal angle \(\theta\in[0,\pi]\) such that \(\cos\theta=x\). Therefore \(\cos(\cos^{-1}x)=x\) for all \(x\in[-1,1]\). Option B (|x|) is incorrect because \(x\) may be negative and composition does not take absolute value. Option C (\(\cos^{-1}x\)) is an angle, not the original numeric value \(x\). Option D (\(-x\)) is also incorrect. Exam tip: remember domain/range restrictions — \(\cos(\cos^{-1}x)=x\) on \([-1,1]\), whereas \(\cos^{-1}(\cos y)\) equals \(y\) only when \(y\in[0,\pi]\).

\(\tan^{-1}x\) denotes the angle whose tangent is x. If \(\tan^{-1}x=0\), taking tan of both sides gives \(x=\tan0=0\). Also note arctan's principal value lies in \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\), so the solution is unique. Option B (1) would be correct only if \(\tan^{-1}x=\tfrac{\pi}{4}\). Option D is an angle (\(\tfrac{\pi}{4}\)), not the value of x. Exam tip: clearly distinguish the inverse-trig function's output (an angle) from the variable x before applying the direct trig function.

The governing concept is the principal-value identity for inverse tangent and inverse cotangent. For a positive real number x, let θ = tan⁻¹x. Then θ lies in (0, π/2), and cot⁻¹x is the complementary angle π/2 − θ under the standard Class 12 principal-value convention. Therefore tan⁻¹x + cot⁻¹x = θ + (π/2 − θ) = π/2. Hence option B is correct. Option A would occur only for a particular value such as x = 1, not for every positive x. Options C and D ignore the complementary principal-value relationship. The restriction x > 0 also avoids sign-related branch complications.

The governing identity for the principal inverse-cosine branch is cos⁻¹(−x) = π − cos⁻¹x for x ∈ [−1, 1]. To see this, put α = cos⁻¹x, so α lies in [0, π] and cos α = x. Since cos(π − α) = −cos α = −x, and π − α is also in the principal range [0, π], it follows that cos⁻¹(−x) = π − α = π − cos⁻¹x. Therefore option C is correct. Option A wrongly applies the odd-function rule; arccos is not odd. Options B and D do not generally produce an angle whose cosine is −x.

If \(x\in\left[-1,1\right]\), then \(\cos(\cos^{-1}x)=x\). Here \(x=-\frac{3}{4}\) lies in the domain.

Here \(\theta\in\left[0,\pi\right]\) and \(\cos\theta=\frac{8}{17}\), so \(\theta\) is in the first quadrant and \(\sin\theta=\frac{15}{17}\). The triangle method is fast.

From \(\tan\theta=\frac{7}{24}\), the hypotenuse is \(\sqrt{7^2+24^2}=25\), so \(\sin\theta=\frac{7}{25}\). Build a triangle from the ratio.

Let \(\theta=\sin^{-1}\frac{4}{5}\), then \(\sin\theta=\frac{4}{5}\) and \(\cos\theta=\frac{3}{5}\). Therefore \(\tan\theta=\frac{4}{3}\).

If \(\theta=\tan^{-1}\frac{12}{5}\), then \(\tan\theta=\frac{12}{5}\) and the hypotenuse is (13). Hence \(\sin\theta=\frac{12}{13}\).

If \(\tan\theta=\frac{15}{8}\), then the hypotenuse is (17). In the principal range \(\theta\) is in the first quadrant, so \(\cos\theta=\frac{8}{17}\).

For every \(x\in\left[-1,1\right]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{2}{3}\) is valid.

From \(\cot\theta=\frac{3}{4}\), the base is (3), perpendicular is (4), and hypotenuse is (5). Hence \(\cos\theta=\frac{3}{5}\).

By the formula, the sum is \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\). While adding \(\tan^{-1}\) terms, check \(ab<1\).

Let \(\theta=\cos^{-1}\frac{5}{13}\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Therefore \(\tan\theta=\frac{12}{5}\).

The inverse tangent function is an odd function. For an odd function, changing the sign of the input changes the sign of the output, so \(f(-x)=-f(x)\). Applying this property to the inverse tangent function gives \(\tan^{-1}(-x)=-\tan^{-1}(x)\). Therefore option A is correct.

The principal value of \(\tan^{-1}(x)\) lies in \((-\frac{\pi}{2},\frac{\pi}{2})\), and the negative of a principal value remains in the same interval. This confirms that no extra \(\pi\) or \(\frac{\pi}{2}\) term is needed. Option B would incorrectly ignore the negative input, while options C and D are not the odd-function identity and generally give a different value. The identity follows because tangent itself is odd and its inverse preserves the corresponding sign relationship on its principal domain.

\(\cos^{-1}(-x)=\pi-\cos^{-1}x\) because the output remains in \([0,\pi]\). Do not treat it as an odd function.

If \(\theta=\cos^{-1}x\), then \(\theta\in[0,\pi]\) and \(\sin\theta\ge0\). Hence the value is \(\sqrt{1-x^2}\).

If (\theta=\sin^{-1}x), then (\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]) and (\cos\theta\ge0). So the positive root should be taken.